Proof of Rules for Complex Derivatives: Carathéodory's Criterion, Continuity, Restriction, Constants, Sums, Products, Quotients, the Chain Rule and Powers

Throughout, routine facts about complex numbers and maps are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); in particular, a function given by a formula exists and takes the stated values, and two maps with the same values are equal (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v), Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)). The continuity results cited below that are stated for the earlier definitions of metric spaces and continuity are read in the current ones by The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (i) and The Earlier and the New Versions of the Definitions of Metric Spaces, Convergence, Continuity and Compactness Agree (vii).

The plan is the one of Carathéodory's approach to the derivative: we first prove (i), and every later rule is then obtained by writing down the function $\varphi$ that the criterion asks for. When $f$ and $g$ are complex differentiable at $z$, we write $\varphi_{f},\varphi_{g}:U\to\mathbb{C}$ for the functions that (i) provides for them, so that $\varphi_{f}$ and $\varphi_{g}$ are continuous at $z$ relative to $U$, $\varphi_{f}(z)=f'(z)$, $\varphi_{g}(z)=g'(z)$, and

$$f(w)-f(z)=\varphi_{f}(w)\,(w-z),\qquad g(w)-g(z)=\varphi_{g}(w)\,(w-z)\qquad(w\in U).$$

Carathéodory's criterion, necessity. Suppose that $f$ is complex differentiable at $z$ with $f'(z)=L$, so that $L$ is the unique complex derivative of $f$ at $z$ (The Complex Derivative at a Point (ii)). Define $\varphi:U\to\mathbb{C}$ by cases (Maps Defined by Cases (i)):

$$\varphi(w)=\frac{f(w)-f(z)}{w-z}\quad(w\in U,\ w\neq z),\qquad \varphi(z)=L .$$

Then $\varphi(z)=L$, and the identity $f(w)-f(z)=\varphi(w)\,(w-z)$ holds for $w\neq z$ by multiplying out, and for $w=z$ because both sides are $0$. It remains to see that $\varphi$ is continuous at $z$ relative to $U$. Distances in $\mathbb{C}$ are moduli of differences (Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (v)). Let $\varepsilon>0$, and choose $\delta>0$ for this $\varepsilon$ as in The Complex Derivative at a Point (i). Let $w\in U$ with $|w-z|<\delta$. If $w\neq z$, then $|\varphi(w)-\varphi(z)|$ is exactly the modulus of the difference quotient minus $L$ bounded there, so it is less than $\varepsilon$; if $w=z$, it is $0$. Hence this $\delta$ serves for $\varepsilon$ in Continuous Map Between Metric Spaces (i).

Carathéodory's criterion, sufficiency. Conversely, let $\varphi:U\to\mathbb{C}$ be continuous at $z$ relative to $U$, with $\varphi(z)=L$ and $f(w)-f(z)=\varphi(w)\,(w-z)$ for $w\in U$. For $w\in U$ with $w\neq z$ we may divide by $w-z$ and obtain

$$\frac{f(w)-f(z)}{w-z}-L=\varphi(w)-\varphi(z).$$

Let $\varepsilon>0$, and choose $\delta>0$ for this $\varepsilon$ by the continuity of $\varphi$ at $z$. Then every $w\in U$ with $0<|w-z|<\delta$ satisfies $|\varphi(w)-\varphi(z)|<\varepsilon$, which is the inequality of The Complex Derivative at a Point (i). So $L$ is a complex derivative of $f$ at $z$; thus $f$ is complex differentiable at $z$, and $f'(z)=L$ by The Complex Derivative at a Point (ii).

Continuity. Let $f$ be complex differentiable at $z$. By (i),

$$f(w)=f(z)+\varphi_{f}(w)\,(w-z)\qquad(w\in U).$$

The right-hand side is built from constant functions, the identity $w\mapsto w$ and $\varphi_{f}$ by sums and products. The constants and the identity are continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (ii), and so is $\varphi_{f}$; hence $f$ is continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (iii).

Restriction. Let $W$ be as in (iii) and let $f$ be complex differentiable at $z$. The restrictions $f|_{W}$ and $\varphi_{f}|_{W}$ are functions $W\to\mathbb{C}$ taking the values of $f$ and of $\varphi_{f}$ (Domains and Restrictions of Maps (ii), Domains and Restrictions of Maps (iii)). Since $W\subseteq U$, the identity of (i) gives

$$f|_{W}(w)-f|_{W}(z)=\varphi_{f}|_{W}(w)\,(w-z)\qquad(w\in W),$$

and $\varphi_{f}|_{W}(z)=f'(z)$. Moreover $\varphi_{f}|_{W}$ is continuous at $z$ relative to $W$ by Restricting a Continuous Map to a Smaller Subset Preserves Continuity (i). Applying (i) to $f|_{W}$ on the open set $W$ in place of $U$, we conclude that $f|_{W}$ is complex differentiable at $z$ with $(f|_{W})'(z)=f'(z)$.

Constants and the identity. If $f$ is constant with value $c$, then $f(w)-f(z)=0=0\cdot(w-z)$, so the constant function $0$ serves as $\varphi$ in (i). If $f(w)=w$, then $f(w)-f(z)=1\cdot(w-z)$, so the constant function $1$ serves. Both are continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (ii), and their values at $z$ are $0$ and $1$, so the derivatives are $0$ and $1$.

Sums and constant multiples. Let $f$ and $g$ be complex differentiable at $z$. For $w\in U$,

$$(f+g)(w)-(f+g)(z)=\bigl(\varphi_{f}(w)+\varphi_{g}(w)\bigr)(w-z),\qquad (\lambda f)(w)-(\lambda f)(z)=\lambda\varphi_{f}(w)\,(w-z).$$

The functions $\varphi_{f}+\varphi_{g}$ and $\lambda\varphi_{f}$ are continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (iii), with values $f'(z)+g'(z)$ and $\lambda f'(z)$ at $z$. By (i), $f+g$ and $\lambda f$ are complex differentiable at $z$, with

$$(f+g)'(z)=f'(z)+g'(z),\qquad (\lambda f)'(z)=\lambda f'(z).$$

Products. Let $f$ and $g$ be complex differentiable at $z$. For $w\in U$, writing $f(w)g(w)-f(z)g(z)=\bigl(f(w)-f(z)\bigr)g(w)+f(z)\bigl(g(w)-g(z)\bigr)$ and inserting the identities for $\varphi_{f}$ and $\varphi_{g}$ gives

$$f(w)g(w)-f(z)g(z)=\bigl(\varphi_{f}(w)\,g(w)+f(z)\,\varphi_{g}(w)\bigr)(w-z).$$

The function $w\mapsto\varphi_{f}(w)\,g(w)+f(z)\,\varphi_{g}(w)$ is continuous at $z$ relative to $U$: $g$ is continuous there by (ii), the constant $f(z)$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (ii), and sums and products of continuous functions are continuous by Continuity of Complex-Valued Functions, and the Modulus of a Power (iii). Its value at $z$ is $f'(z)g(z)+f(z)g'(z)$. By (i), $fg$ is complex differentiable at $z$ and

$$(fg)'(z)=f'(z)\,g(z)+f(z)\,g'(z).$$

Reciprocals. Let $g$ be complex differentiable at $z$, with $g(w)\neq0$ for every $w\in U$. For $w\in U$,

$$\frac{1}{g(w)}-\frac{1}{g(z)}=\frac{g(z)-g(w)}{g(w)\,g(z)}=-\frac{\varphi_{g}(w)}{g(w)\,g(z)}\,(w-z).$$

The coefficient of $w-z$ is the product of $\varphi_{g}$, of $1/g$ and of the constant $-1/g(z)$. Here $g$ is continuous at $z$ by (ii), hence so is $1/g$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (iv); the constant is continuous by Continuity of Complex-Valued Functions, and the Modulus of a Power (ii), and so the product is continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (iii). Its value at $z$ is $-g'(z)/g(z)^{2}$, so by (i) the function $1/g$ is complex differentiable at $z$ and

$$\Bigl(\frac{1}{g}\Bigr)'(z)=-\frac{g'(z)}{g(z)^{2}}.$$

Quotients. Let $f$ and $g$ be complex differentiable at $z$, with $g$ nowhere $0$ on $U$. The functions $f/g$ and $f\cdot(1/g)$ have the same values, so they are equal. By (ix), $1/g$ is complex differentiable at $z$, and then by (viii) so is $f/g$, with

$$\Bigl(\frac{f}{g}\Bigr)'(z)=f'(z)\,\frac{1}{g(z)}+f(z)\Bigl(-\frac{g'(z)}{g(z)^{2}}\Bigr)=\frac{f'(z)\,g(z)-f(z)\,g'(z)}{g(z)^{2}}.$$

The chain rule. Let $V$ and $h$ be as in (xi), with $f$ complex differentiable at $z$ and $h$ complex differentiable at $f(z)$. Let $\psi:V\to\mathbb{C}$ be the function that (i) provides for $h$ at $f(z)$: it is continuous at $f(z)$ relative to $V$, $\psi(f(z))=h'(f(z))$, and $h(v)-h(f(z))=\psi(v)\,(v-f(z))$ for $v\in V$. Taking $v=f(w)$ and then using the identity for $\varphi_{f}$, we get for $w\in U$

$$h(f(w))-h(f(z))=\psi(f(w))\,\bigl(f(w)-f(z)\bigr)=\psi(f(w))\,\varphi_{f}(w)\,(w-z).$$

Since $f$ is continuous at $z$ relative to $U$ by (ii), the composite $\psi\circ f$ is continuous at $z$ relative to $U$ by Continuity of Complex-Valued Functions, and the Modulus of a Power (v), and hence so is the coefficient $(\psi\circ f)\,\varphi_{f}$, by Continuity of Complex-Valued Functions, and the Modulus of a Power (iii). Its value at $z$ is $h'(f(z))\,f'(z)$. By (i), $h\circ f$ is complex differentiable at $z$ and

$$(h\circ f)'(z)=h'\bigl(f(z)\bigr)\,f'(z).$$

Powers. For $n\in\mathbb{N}_{0}$ let $f_{n}:U\to\mathbb{C}$ be the function $f_{n}(w)=w^{n+1}$. We prove by induction on $n$ that $f_{n}$ is complex differentiable at $z$ with

$$f_{n}'(z)=(n+1)\,z^{n}.$$

For $n=0$ we have $w^{1}=w$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)), so $f_{0}$ is the identity, which by (v) has derivative $1=1\cdot z^{0}$ at $z$ (Powers with Exponents in the Natural Numbers with Zero (ii)). Suppose the claim holds for $n$. Since $w^{n+2}=w^{n+1}\,w$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (i)), $f_{n+1}$ is the product of $f_{n}$ and the identity, and (viii) together with the induction hypothesis and (v) shows that it is complex differentiable at $z$ with derivative

$$(n+1)\,z^{n}\cdot z+z^{n+1}\cdot1=(n+2)\,z^{n+1}.$$

This completes the induction.